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Effective descent for modules along faithfully flat ring maps.

Proposition 35.3.9. Let $R \to A$ be a faithfully flat ring map. Then

  1. any descent datum on modules with respect to $R \to A$ is effective,

  2. the functor $M \mapsto (A \otimes _ R M, can)$ from $R$-modules to the category of descent data is an equivalence, and

  3. the inverse functor is given by $(N, \varphi ) \mapsto H^0(s(N_\bullet ))$.

Proof. We only prove (1) and omit the proofs of (2) and (3). As $R \to A$ is faithfully flat, there exists a faithfully flat base change $R \to R'$ such that $R' \to A' = R' \otimes _ R A$ has a section (namely take $R' = A$ as in the proof of Lemma 35.3.6). Hence, using Lemma 35.3.8 we may assume that $R \to A$ has a section, say $\sigma : A \to R$. Let $(N, \varphi )$ be a descent datum relative to $R \to A$. Set

\[ M = H^0(s(N_\bullet )) = \{ n \in N \mid 1 \otimes n = \varphi (n \otimes 1)\} \subset N \]

By Lemma 35.3.7 it suffices to show that $A \otimes _ R M \to N$ is an isomorphism.

Take an element $n \in N$. Write $\varphi (n \otimes 1) = \sum a_ i \otimes x_ i$ for certain $a_ i \in A$ and $x_ i \in N$. By Lemma 35.3.2 we have $n = \sum a_ i x_ i$ in $N$ (because $\sigma ^0_0 \circ \delta ^1_1 = \text{id}$ in any cosimplicial object). Next, write $\varphi (x_ i \otimes 1) = \sum a_{ij} \otimes y_ j$ for certain $a_{ij} \in A$ and $y_ j \in N$. The cocycle condition means that

\[ \sum a_ i \otimes a_{ij} \otimes y_ j = \sum a_ i \otimes 1 \otimes x_ i \]

in $A \otimes _ R A \otimes _ R N$. We conclude two things from this:

  1. applying $\sigma $ to the first $A$ we get $\sum \sigma (a_ i) \varphi (x_ i \otimes 1) = \sum \sigma (a_ i) \otimes x_ i$,

  2. applying $\sigma $ to the middle $A$ we get $\sum _ i a_ i \otimes \sum _ j \sigma (a_{ij}) y_ j = \sum a_ i \otimes x_ i$.

Part (1) shows that $\sum \sigma (a_ i) x_ i \in M$. Applying this to $x_ i$ we see that $\sum \sigma (a_{ij})y_ i \in M$ for all $i$. Multiplying out the equation in (2) we conclude that $\sum _ i a_ i (\sum _ j \sigma (a_{ij}) y_ j) = \sum a_ i x_ i = n$. Hence $A \otimes _ R M \to N$ is surjective. Finally, suppose that $m_ i \in M$ and $\sum a_ i m_ i = 0$. Then we see by applying $\varphi $ to $\sum a_ im_ i \otimes 1$ that $\sum a_ i \otimes m_ i = 0$. In other words $A \otimes _ R M \to N$ is injective and we win. $\square$

Comments (10)

Comment #874 by Bhargav Bhatt on

Suggested slogan: One has effective descent for modules along faithfully flat ring maps.

Comment #2832 by Ko Aoki on

Two typos in the proof: "we may assume that as a section" should be replaced by "we may assume that has a section", and "(because in any cosimplicial object)" should be replaced by "(because in any cosimplicial object)"

Comment #4067 by Stephan on

Could you elaborate on the "Hence by the first conclusion we see that..." part? why is in M?

Comment #4144 by on

The first conclusion was that is in . Apply this to to get in . OK?

Comment #7526 by WhatJiaranEatsTonight on

Could you please tell me how to deduce that is in the image of ? I cannot see it from the fact that and .

Comment #7529 by on

Because is in by the first conclusion applied to . But I think we need to rewrite the proof as it is too succint.

Comment #8682 by Figo on

I am sorry that I cannot get from the fact . I know that from the conclusion by (1). And as , I only can conclude that .

Comment #8683 by Figo on

Oh I am sorry. "Apply to x_inx_i$ and use the same argument.

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  • 4 comment(s) on Section 35.3: Descent for modules

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